Mixing
Any thoughts on this integral?
$begingroup$
$int cos^2(x)cdotsin^4(x)dx$
I tried the usual trigonometric identities but they don't seem helpful
integration indefinite-integrals
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
asked Jan 23 at 10:55
ChangaChanga
234
$endgroup$
add a comment |
$begingroup$
$int cos^2(x)cdotsin^4(x)dx$
I tried the usual trigonometric identities but they don't seem helpful
integration indefinite-integrals
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
asked Jan 23 at 10:55
ChangaChanga
234
$endgroup$
1
$begingroup$
As both exponents are even, you have to linearise. The simplest way uses the complex exponential.
$endgroup$
– Bernard
Jan 23 at 11:01
$begingroup$
You could try with $cos^2=1-sin^2$, then decompose $sin^4=sin^2cdotsin^2$, and similarly forma $sin^6$. Working this way you should lower the exponent by partial integration
$endgroup$
– Gabriele Cassese
Jan 23 at 11:01
$begingroup$
You can try t=sin^3(x)
$endgroup$
– Shaq
Jan 23 at 11:02
add a comment |
$begingroup$
$int cos^2(x)cdotsin^4(x)dx$
I tried the usual trigonometric identities but they don't seem helpful
integration indefinite-integrals
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
asked Jan 23 at 10:55
ChangaChanga
234
$endgroup$
$int cos^2(x)cdotsin^4(x)dx$
I tried the usual trigonometric identities but they don't seem helpful
integration indefinite-integrals
integration indefinite-integrals
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
asked Jan 23 at 10:55
ChangaChanga
234
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
asked Jan 23 at 10:55
ChangaChanga
234
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
edited Jan 23 at 10:59
Gabriele Cassese
1,021314
1,021314
asked Jan 23 at 10:55
ChangaChanga
234
asked Jan 23 at 10:55
ChangaChanga
234
asked Jan 23 at 10:55
ChangaChanga
234
234
1
$begingroup$
As both exponents are even, you have to linearise. The simplest way uses the complex exponential.
$endgroup$
– Bernard
Jan 23 at 11:01
$begingroup$
You could try with $cos^2=1-sin^2$, then decompose $sin^4=sin^2cdotsin^2$, and similarly forma $sin^6$. Working this way you should lower the exponent by partial integration
$endgroup$
– Gabriele Cassese
Jan 23 at 11:01
$begingroup$
You can try t=sin^3(x)
$endgroup$
– Shaq
Jan 23 at 11:02
add a comment |
1
$begingroup$
As both exponents are even, you have to linearise. The simplest way uses the complex exponential.
$endgroup$
– Bernard
Jan 23 at 11:01
$begingroup$
You could try with $cos^2=1-sin^2$, then decompose $sin^4=sin^2cdotsin^2$, and similarly forma $sin^6$. Working this way you should lower the exponent by partial integration
$endgroup$
– Gabriele Cassese
Jan 23 at 11:01
$begingroup$
You can try t=sin^3(x)
$endgroup$
– Shaq
Jan 23 at 11:02
1
1
$begingroup$
As both exponents are even, you have to linearise. The simplest way uses the complex exponential.
$endgroup$
– Bernard
Jan 23 at 11:01
$begingroup$
As both exponents are even, you have to linearise. The simplest way uses the complex exponential.
$endgroup$
– Bernard
Jan 23 at 11:01
$begingroup$
You could try with $cos^2=1-sin^2$, then decompose $sin^4=sin^2cdotsin^2$, and similarly forma $sin^6$. Working this way you should lower the exponent by partial integration
$endgroup$
– Gabriele Cassese
Jan 23 at 11:01
$begingroup$
You could try with $cos^2=1-sin^2$, then decompose $sin^4=sin^2cdotsin^2$, and similarly forma $sin^6$. Working this way you should lower the exponent by partial integration
$endgroup$
– Gabriele Cassese
Jan 23 at 11:01
$begingroup$
You can try t=sin^3(x)
$endgroup$
– Shaq
Jan 23 at 11:02
$begingroup$
You can try t=sin^3(x)
$endgroup$
– Shaq
Jan 23 at 11:02
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
$intcos^2(x)sin^4(x)~dx=frac14intsin^2(2x)sin^2(x)~dx\=frac18intsin^2(2x)[1-cos(2x)]~dx\=frac18left[intsin^2(2x)~dx-intsin^2(2x)cos(2x)~dxright]$
Solve the first integral by writing $sin^2(2x)=frac12[1-cos(4x)]$ and the second one by putting $sin(2x)=t$.
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thank you very much! it's clear now
$endgroup$
– Changa
Jan 23 at 11:11
$begingroup$
Glad to help :)
$endgroup$
– Shubham Johri
Jan 23 at 11:11
add a comment |
$begingroup$
Hint:
Use Intuition behind euler's formula
$$(2cos x)^2(2isin x)^4=left(e^{ix}+e^{-ix}right)^2left(e^{ix}-e^{-ix}right)^4$$
If $2cos(nx)=e^{inx}+e^{-inx}=u_n$
$$64cos^2xsin^4x=u_6+u_4(-4+2)+u_2(2+1-8+1+6)+u_0(-4)+12$$
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Use $cos(x)^2 = 1-sin(x)^2$. Then you have a $sin(x)^4$ and a $sin(x)^6$ to integrate. These can either be computed by partial integration or by using trigonometric identities for powers of $sin$.
answered Jan 23 at 11:01
KlausKlaus
2,30712
$endgroup$
add a comment |
$begingroup$
Write $c$ for $cos x$ and $s$ for $sin x$. Then the integrand is
begin{align}
c^2 s^4
&= c^2 s^2 (s^2) \
&= frac{1}{4} (2cs)^2 (1 - c^2) \
&= frac{1}{2} frac{1}{4} (2cs)^2 (2 - 2c^2) \
&= frac{1}{2} left( frac{1}{4} (2cs)^2 (1 - 2c^2) right)
+ frac{1}{2} left( frac{1}{4} (2cs)^2 right) \
end{align}
Now $2cs = sin 2x$, and $1 - 2c^2 = -cos 2x$, so from here things should be relatively simple.
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
$endgroup$
add a comment |
$begingroup$
You can lower the degree by noting that
$$
cos^2x=frac{1+cos2x}{2},qquad sin^2x=frac{1-cos2x}{2}
$$
Thus you get
$$
frac{(1-cos^22x)(1-cos2x)}{4}=frac{1}{4}(1-cos^22x-cos2x+cos^32x)=
frac{1}{4}left(1-frac{1+cos4x}{2}-cos2xsin^22xright)
$$
answered Jan 23 at 11:12
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
Here's a cool thing. It's called a reduction formula.
Consider the integral
$$I(n)=int cos(x)^{n}sin(x)^{2n}mathrm dx$$
Then recall that $sin(x)^2=1-cos(x)^2$:
$$I(n)=int cos(x)^nleft(1-cos(x)^2right)^nmathrm dx$$
Then assuming that $n$ is a non-negative integer, we recall the binomial formula:
$$(a-b)^n=sum_{k=0}^{n}(-1)^k{nchoose k}a^{n-k}b^k$$
$$left(1-cos(x)^2right)^n=sum_{k=0}^{n}(-1)^k{nchoose k}cos(x)^{2k}$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}intcos(x)^{2k+n}mathrm dx$$
Then consider the integral
$$C(m)=intcos(x)^mmathrm dx$$
$$C(m)=intcos(x)^{m-1}cos(x)mathrm dx$$
IBP:
$$mathrm dv=cos(x)mathrm dxRightarrow v=sin(x)$$
$$u=cos(x)^{m-1}Rightarrow mathrm du=-(m-1)cos(x)^{m-2}sin(x)mathrm dx$$
So
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}sin(x)^2mathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}mathrm dx-(m-1)intcos(x)^mmathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)-(m-1)C(m)$$
$$mC(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)$$
$$C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}C(2k+n)$$
I know this isn't a very efficient method for too large to count on one hand, but it is still an alternate method. Because your integral is $I(2)$, we have that
$$I(2)=sum_{k=0}^{2}(-1)^k{2choose k}C(2k+2)$$
$$I(2)=C(2)-2C(4)+C(6)$$
And from $C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$ we see that
$$I(2)=C(2)-2left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}C(4)$$
$$I(2)=-frac12C(2)-frac{cos(x)^{3}sin(x)}{2}+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]$$
$$I(2)=frac18C(2)-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And since $C(0)=intmathrm dx=x$,
$$I(2)=frac18left[frac{cos(x)sin(x)}{2}+frac{x}{2}right]-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
$$I(2)=frac{x}{16}+frac{cos(x)sin(x)}{16}-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And there you go.
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
$endgroup$
add a comment |
$begingroup$
You could also use Euler’s Formula.
$$ sin x = frac{e^{ix} - e^{-ix }}{2i} $$
$$ cos x = frac{e^{ix} + e^{-ix}}{2} $$
$$int {{(frac{e^{ix} - e^{-ix }}{2i})}^2} bullet {(frac{e^{ix} - e^{-ix }}{2i})}^4 dx $$
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084326%2fany-thoughts-on-this-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$intcos^2(x)sin^4(x)~dx=frac14intsin^2(2x)sin^2(x)~dx\=frac18intsin^2(2x)[1-cos(2x)]~dx\=frac18left[intsin^2(2x)~dx-intsin^2(2x)cos(2x)~dxright]$
Solve the first integral by writing $sin^2(2x)=frac12[1-cos(4x)]$ and the second one by putting $sin(2x)=t$.
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thank you very much! it's clear now
$endgroup$
– Changa
Jan 23 at 11:11
$begingroup$
Glad to help :)
$endgroup$
– Shubham Johri
Jan 23 at 11:11
add a comment |
$begingroup$
$intcos^2(x)sin^4(x)~dx=frac14intsin^2(2x)sin^2(x)~dx\=frac18intsin^2(2x)[1-cos(2x)]~dx\=frac18left[intsin^2(2x)~dx-intsin^2(2x)cos(2x)~dxright]$
Solve the first integral by writing $sin^2(2x)=frac12[1-cos(4x)]$ and the second one by putting $sin(2x)=t$.
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
$endgroup$
$begingroup$
Thank you very much! it's clear now
$endgroup$
– Changa
Jan 23 at 11:11
$begingroup$
Glad to help :)
$endgroup$
– Shubham Johri
Jan 23 at 11:11
add a comment |
$begingroup$
$intcos^2(x)sin^4(x)~dx=frac14intsin^2(2x)sin^2(x)~dx\=frac18intsin^2(2x)[1-cos(2x)]~dx\=frac18left[intsin^2(2x)~dx-intsin^2(2x)cos(2x)~dxright]$
Solve the first integral by writing $sin^2(2x)=frac12[1-cos(4x)]$ and the second one by putting $sin(2x)=t$.
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
$endgroup$
$intcos^2(x)sin^4(x)~dx=frac14intsin^2(2x)sin^2(x)~dx\=frac18intsin^2(2x)[1-cos(2x)]~dx\=frac18left[intsin^2(2x)~dx-intsin^2(2x)cos(2x)~dxright]$
Solve the first integral by writing $sin^2(2x)=frac12[1-cos(4x)]$ and the second one by putting $sin(2x)=t$.
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
answered Jan 23 at 11:09
Shubham JohriShubham Johri
5,204718
5,204718
$begingroup$
Thank you very much! it's clear now
$endgroup$
– Changa
Jan 23 at 11:11
$begingroup$
Glad to help :)
$endgroup$
– Shubham Johri
Jan 23 at 11:11
add a comment |
$begingroup$
Thank you very much! it's clear now
$endgroup$
– Changa
Jan 23 at 11:11
$begingroup$
Glad to help :)
$endgroup$
– Shubham Johri
Jan 23 at 11:11
$begingroup$
Thank you very much! it's clear now
$endgroup$
– Changa
Jan 23 at 11:11
$begingroup$
Thank you very much! it's clear now
$endgroup$
– Changa
Jan 23 at 11:11
$begingroup$
Glad to help :)
$endgroup$
– Shubham Johri
Jan 23 at 11:11
$begingroup$
Glad to help :)
$endgroup$
– Shubham Johri
Jan 23 at 11:11
add a comment |
$begingroup$
Hint:
Use Intuition behind euler's formula
$$(2cos x)^2(2isin x)^4=left(e^{ix}+e^{-ix}right)^2left(e^{ix}-e^{-ix}right)^4$$
If $2cos(nx)=e^{inx}+e^{-inx}=u_n$
$$64cos^2xsin^4x=u_6+u_4(-4+2)+u_2(2+1-8+1+6)+u_0(-4)+12$$
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Use Intuition behind euler's formula
$$(2cos x)^2(2isin x)^4=left(e^{ix}+e^{-ix}right)^2left(e^{ix}-e^{-ix}right)^4$$
If $2cos(nx)=e^{inx}+e^{-inx}=u_n$
$$64cos^2xsin^4x=u_6+u_4(-4+2)+u_2(2+1-8+1+6)+u_0(-4)+12$$
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
add a comment |
$begingroup$
Hint:
Use Intuition behind euler's formula
$$(2cos x)^2(2isin x)^4=left(e^{ix}+e^{-ix}right)^2left(e^{ix}-e^{-ix}right)^4$$
If $2cos(nx)=e^{inx}+e^{-inx}=u_n$
$$64cos^2xsin^4x=u_6+u_4(-4+2)+u_2(2+1-8+1+6)+u_0(-4)+12$$
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
$endgroup$
Hint:
Use Intuition behind euler's formula
$$(2cos x)^2(2isin x)^4=left(e^{ix}+e^{-ix}right)^2left(e^{ix}-e^{-ix}right)^4$$
If $2cos(nx)=e^{inx}+e^{-inx}=u_n$
$$64cos^2xsin^4x=u_6+u_4(-4+2)+u_2(2+1-8+1+6)+u_0(-4)+12$$
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
answered Jan 23 at 11:46
lab bhattacharjeelab bhattacharjee
226k15158275
226k15158275
add a comment |
add a comment |
$begingroup$
Use $cos(x)^2 = 1-sin(x)^2$. Then you have a $sin(x)^4$ and a $sin(x)^6$ to integrate. These can either be computed by partial integration or by using trigonometric identities for powers of $sin$.
answered Jan 23 at 11:01
KlausKlaus
2,30712
$endgroup$
add a comment |
$begingroup$
Use $cos(x)^2 = 1-sin(x)^2$. Then you have a $sin(x)^4$ and a $sin(x)^6$ to integrate. These can either be computed by partial integration or by using trigonometric identities for powers of $sin$.
answered Jan 23 at 11:01
KlausKlaus
2,30712
$endgroup$
add a comment |
$begingroup$
Use $cos(x)^2 = 1-sin(x)^2$. Then you have a $sin(x)^4$ and a $sin(x)^6$ to integrate. These can either be computed by partial integration or by using trigonometric identities for powers of $sin$.
answered Jan 23 at 11:01
KlausKlaus
2,30712
$endgroup$
Use $cos(x)^2 = 1-sin(x)^2$. Then you have a $sin(x)^4$ and a $sin(x)^6$ to integrate. These can either be computed by partial integration or by using trigonometric identities for powers of $sin$.
answered Jan 23 at 11:01
KlausKlaus
2,30712
answered Jan 23 at 11:01
KlausKlaus
2,30712
answered Jan 23 at 11:01
KlausKlaus
2,30712
answered Jan 23 at 11:01
KlausKlaus
2,30712
2,30712
add a comment |
add a comment |
$begingroup$
Write $c$ for $cos x$ and $s$ for $sin x$. Then the integrand is
begin{align}
c^2 s^4
&= c^2 s^2 (s^2) \
&= frac{1}{4} (2cs)^2 (1 - c^2) \
&= frac{1}{2} frac{1}{4} (2cs)^2 (2 - 2c^2) \
&= frac{1}{2} left( frac{1}{4} (2cs)^2 (1 - 2c^2) right)
+ frac{1}{2} left( frac{1}{4} (2cs)^2 right) \
end{align}
Now $2cs = sin 2x$, and $1 - 2c^2 = -cos 2x$, so from here things should be relatively simple.
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
$endgroup$
add a comment |
$begingroup$
Write $c$ for $cos x$ and $s$ for $sin x$. Then the integrand is
begin{align}
c^2 s^4
&= c^2 s^2 (s^2) \
&= frac{1}{4} (2cs)^2 (1 - c^2) \
&= frac{1}{2} frac{1}{4} (2cs)^2 (2 - 2c^2) \
&= frac{1}{2} left( frac{1}{4} (2cs)^2 (1 - 2c^2) right)
+ frac{1}{2} left( frac{1}{4} (2cs)^2 right) \
end{align}
Now $2cs = sin 2x$, and $1 - 2c^2 = -cos 2x$, so from here things should be relatively simple.
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
$endgroup$
add a comment |
$begingroup$
Write $c$ for $cos x$ and $s$ for $sin x$. Then the integrand is
begin{align}
c^2 s^4
&= c^2 s^2 (s^2) \
&= frac{1}{4} (2cs)^2 (1 - c^2) \
&= frac{1}{2} frac{1}{4} (2cs)^2 (2 - 2c^2) \
&= frac{1}{2} left( frac{1}{4} (2cs)^2 (1 - 2c^2) right)
+ frac{1}{2} left( frac{1}{4} (2cs)^2 right) \
end{align}
Now $2cs = sin 2x$, and $1 - 2c^2 = -cos 2x$, so from here things should be relatively simple.
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
$endgroup$
Write $c$ for $cos x$ and $s$ for $sin x$. Then the integrand is
begin{align}
c^2 s^4
&= c^2 s^2 (s^2) \
&= frac{1}{4} (2cs)^2 (1 - c^2) \
&= frac{1}{2} frac{1}{4} (2cs)^2 (2 - 2c^2) \
&= frac{1}{2} left( frac{1}{4} (2cs)^2 (1 - 2c^2) right)
+ frac{1}{2} left( frac{1}{4} (2cs)^2 right) \
end{align}
Now $2cs = sin 2x$, and $1 - 2c^2 = -cos 2x$, so from here things should be relatively simple.
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
answered Jan 23 at 11:06
John HughesJohn Hughes
64.5k24191
64.5k24191
add a comment |
add a comment |
$begingroup$
You can lower the degree by noting that
$$
cos^2x=frac{1+cos2x}{2},qquad sin^2x=frac{1-cos2x}{2}
$$
Thus you get
$$
frac{(1-cos^22x)(1-cos2x)}{4}=frac{1}{4}(1-cos^22x-cos2x+cos^32x)=
frac{1}{4}left(1-frac{1+cos4x}{2}-cos2xsin^22xright)
$$
answered Jan 23 at 11:12
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You can lower the degree by noting that
$$
cos^2x=frac{1+cos2x}{2},qquad sin^2x=frac{1-cos2x}{2}
$$
Thus you get
$$
frac{(1-cos^22x)(1-cos2x)}{4}=frac{1}{4}(1-cos^22x-cos2x+cos^32x)=
frac{1}{4}left(1-frac{1+cos4x}{2}-cos2xsin^22xright)
$$
answered Jan 23 at 11:12
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You can lower the degree by noting that
$$
cos^2x=frac{1+cos2x}{2},qquad sin^2x=frac{1-cos2x}{2}
$$
Thus you get
$$
frac{(1-cos^22x)(1-cos2x)}{4}=frac{1}{4}(1-cos^22x-cos2x+cos^32x)=
frac{1}{4}left(1-frac{1+cos4x}{2}-cos2xsin^22xright)
$$
answered Jan 23 at 11:12
egregegreg
184k1486205
$endgroup$
You can lower the degree by noting that
$$
cos^2x=frac{1+cos2x}{2},qquad sin^2x=frac{1-cos2x}{2}
$$
Thus you get
$$
frac{(1-cos^22x)(1-cos2x)}{4}=frac{1}{4}(1-cos^22x-cos2x+cos^32x)=
frac{1}{4}left(1-frac{1+cos4x}{2}-cos2xsin^22xright)
$$
answered Jan 23 at 11:12
egregegreg
184k1486205
answered Jan 23 at 11:12
egregegreg
184k1486205
answered Jan 23 at 11:12
egregegreg
184k1486205
answered Jan 23 at 11:12
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
$begingroup$
Here's a cool thing. It's called a reduction formula.
Consider the integral
$$I(n)=int cos(x)^{n}sin(x)^{2n}mathrm dx$$
Then recall that $sin(x)^2=1-cos(x)^2$:
$$I(n)=int cos(x)^nleft(1-cos(x)^2right)^nmathrm dx$$
Then assuming that $n$ is a non-negative integer, we recall the binomial formula:
$$(a-b)^n=sum_{k=0}^{n}(-1)^k{nchoose k}a^{n-k}b^k$$
$$left(1-cos(x)^2right)^n=sum_{k=0}^{n}(-1)^k{nchoose k}cos(x)^{2k}$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}intcos(x)^{2k+n}mathrm dx$$
Then consider the integral
$$C(m)=intcos(x)^mmathrm dx$$
$$C(m)=intcos(x)^{m-1}cos(x)mathrm dx$$
IBP:
$$mathrm dv=cos(x)mathrm dxRightarrow v=sin(x)$$
$$u=cos(x)^{m-1}Rightarrow mathrm du=-(m-1)cos(x)^{m-2}sin(x)mathrm dx$$
So
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}sin(x)^2mathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}mathrm dx-(m-1)intcos(x)^mmathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)-(m-1)C(m)$$
$$mC(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)$$
$$C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}C(2k+n)$$
I know this isn't a very efficient method for too large to count on one hand, but it is still an alternate method. Because your integral is $I(2)$, we have that
$$I(2)=sum_{k=0}^{2}(-1)^k{2choose k}C(2k+2)$$
$$I(2)=C(2)-2C(4)+C(6)$$
And from $C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$ we see that
$$I(2)=C(2)-2left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}C(4)$$
$$I(2)=-frac12C(2)-frac{cos(x)^{3}sin(x)}{2}+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]$$
$$I(2)=frac18C(2)-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And since $C(0)=intmathrm dx=x$,
$$I(2)=frac18left[frac{cos(x)sin(x)}{2}+frac{x}{2}right]-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
$$I(2)=frac{x}{16}+frac{cos(x)sin(x)}{16}-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And there you go.
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
$endgroup$
add a comment |
$begingroup$
Here's a cool thing. It's called a reduction formula.
Consider the integral
$$I(n)=int cos(x)^{n}sin(x)^{2n}mathrm dx$$
Then recall that $sin(x)^2=1-cos(x)^2$:
$$I(n)=int cos(x)^nleft(1-cos(x)^2right)^nmathrm dx$$
Then assuming that $n$ is a non-negative integer, we recall the binomial formula:
$$(a-b)^n=sum_{k=0}^{n}(-1)^k{nchoose k}a^{n-k}b^k$$
$$left(1-cos(x)^2right)^n=sum_{k=0}^{n}(-1)^k{nchoose k}cos(x)^{2k}$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}intcos(x)^{2k+n}mathrm dx$$
Then consider the integral
$$C(m)=intcos(x)^mmathrm dx$$
$$C(m)=intcos(x)^{m-1}cos(x)mathrm dx$$
IBP:
$$mathrm dv=cos(x)mathrm dxRightarrow v=sin(x)$$
$$u=cos(x)^{m-1}Rightarrow mathrm du=-(m-1)cos(x)^{m-2}sin(x)mathrm dx$$
So
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}sin(x)^2mathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}mathrm dx-(m-1)intcos(x)^mmathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)-(m-1)C(m)$$
$$mC(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)$$
$$C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}C(2k+n)$$
I know this isn't a very efficient method for too large to count on one hand, but it is still an alternate method. Because your integral is $I(2)$, we have that
$$I(2)=sum_{k=0}^{2}(-1)^k{2choose k}C(2k+2)$$
$$I(2)=C(2)-2C(4)+C(6)$$
And from $C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$ we see that
$$I(2)=C(2)-2left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}C(4)$$
$$I(2)=-frac12C(2)-frac{cos(x)^{3}sin(x)}{2}+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]$$
$$I(2)=frac18C(2)-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And since $C(0)=intmathrm dx=x$,
$$I(2)=frac18left[frac{cos(x)sin(x)}{2}+frac{x}{2}right]-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
$$I(2)=frac{x}{16}+frac{cos(x)sin(x)}{16}-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And there you go.
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
$endgroup$
add a comment |
$begingroup$
Here's a cool thing. It's called a reduction formula.
Consider the integral
$$I(n)=int cos(x)^{n}sin(x)^{2n}mathrm dx$$
Then recall that $sin(x)^2=1-cos(x)^2$:
$$I(n)=int cos(x)^nleft(1-cos(x)^2right)^nmathrm dx$$
Then assuming that $n$ is a non-negative integer, we recall the binomial formula:
$$(a-b)^n=sum_{k=0}^{n}(-1)^k{nchoose k}a^{n-k}b^k$$
$$left(1-cos(x)^2right)^n=sum_{k=0}^{n}(-1)^k{nchoose k}cos(x)^{2k}$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}intcos(x)^{2k+n}mathrm dx$$
Then consider the integral
$$C(m)=intcos(x)^mmathrm dx$$
$$C(m)=intcos(x)^{m-1}cos(x)mathrm dx$$
IBP:
$$mathrm dv=cos(x)mathrm dxRightarrow v=sin(x)$$
$$u=cos(x)^{m-1}Rightarrow mathrm du=-(m-1)cos(x)^{m-2}sin(x)mathrm dx$$
So
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}sin(x)^2mathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}mathrm dx-(m-1)intcos(x)^mmathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)-(m-1)C(m)$$
$$mC(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)$$
$$C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}C(2k+n)$$
I know this isn't a very efficient method for too large to count on one hand, but it is still an alternate method. Because your integral is $I(2)$, we have that
$$I(2)=sum_{k=0}^{2}(-1)^k{2choose k}C(2k+2)$$
$$I(2)=C(2)-2C(4)+C(6)$$
And from $C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$ we see that
$$I(2)=C(2)-2left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}C(4)$$
$$I(2)=-frac12C(2)-frac{cos(x)^{3}sin(x)}{2}+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]$$
$$I(2)=frac18C(2)-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And since $C(0)=intmathrm dx=x$,
$$I(2)=frac18left[frac{cos(x)sin(x)}{2}+frac{x}{2}right]-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
$$I(2)=frac{x}{16}+frac{cos(x)sin(x)}{16}-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And there you go.
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
$endgroup$
Here's a cool thing. It's called a reduction formula.
Consider the integral
$$I(n)=int cos(x)^{n}sin(x)^{2n}mathrm dx$$
Then recall that $sin(x)^2=1-cos(x)^2$:
$$I(n)=int cos(x)^nleft(1-cos(x)^2right)^nmathrm dx$$
Then assuming that $n$ is a non-negative integer, we recall the binomial formula:
$$(a-b)^n=sum_{k=0}^{n}(-1)^k{nchoose k}a^{n-k}b^k$$
$$left(1-cos(x)^2right)^n=sum_{k=0}^{n}(-1)^k{nchoose k}cos(x)^{2k}$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}intcos(x)^{2k+n}mathrm dx$$
Then consider the integral
$$C(m)=intcos(x)^mmathrm dx$$
$$C(m)=intcos(x)^{m-1}cos(x)mathrm dx$$
IBP:
$$mathrm dv=cos(x)mathrm dxRightarrow v=sin(x)$$
$$u=cos(x)^{m-1}Rightarrow mathrm du=-(m-1)cos(x)^{m-2}sin(x)mathrm dx$$
So
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}sin(x)^2mathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)intcos(x)^{m-2}mathrm dx-(m-1)intcos(x)^mmathrm dx$$
$$C(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)-(m-1)C(m)$$
$$mC(m)=cos(x)^{m-1}sin(x)+(m-1)C(m-2)$$
$$C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$$
So
$$I(n)=sum_{k=0}^{n}(-1)^k{nchoose k}C(2k+n)$$
I know this isn't a very efficient method for too large to count on one hand, but it is still an alternate method. Because your integral is $I(2)$, we have that
$$I(2)=sum_{k=0}^{2}(-1)^k{2choose k}C(2k+2)$$
$$I(2)=C(2)-2C(4)+C(6)$$
And from $C(m)=frac{cos(x)^{m-1}sin(x)}{m}+frac{m-1}{m}C(m-2)$ we see that
$$I(2)=C(2)-2left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}C(4)$$
$$I(2)=-frac12C(2)-frac{cos(x)^{3}sin(x)}{2}+frac{cos(x)^{5}sin(x)}{6}+frac{5}{6}left[frac{cos(x)^{3}sin(x)}{4}+frac{3}{4}C(2)right]$$
$$I(2)=frac18C(2)-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And since $C(0)=intmathrm dx=x$,
$$I(2)=frac18left[frac{cos(x)sin(x)}{2}+frac{x}{2}right]-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
$$I(2)=frac{x}{16}+frac{cos(x)sin(x)}{16}-frac{7cos(x)^{3}sin(x)}{24}+frac{cos(x)^{5}sin(x)}{6}$$
And there you go.
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
answered Jan 24 at 16:59
clathratusclathratus
4,9301338
4,9301338
add a comment |
add a comment |
$begingroup$
You could also use Euler’s Formula.
$$ sin x = frac{e^{ix} - e^{-ix }}{2i} $$
$$ cos x = frac{e^{ix} + e^{-ix}}{2} $$
$$int {{(frac{e^{ix} - e^{-ix }}{2i})}^2} bullet {(frac{e^{ix} - e^{-ix }}{2i})}^4 dx $$
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
$endgroup$
add a comment |
$begingroup$
You could also use Euler’s Formula.
$$ sin x = frac{e^{ix} - e^{-ix }}{2i} $$
$$ cos x = frac{e^{ix} + e^{-ix}}{2} $$
$$int {{(frac{e^{ix} - e^{-ix }}{2i})}^2} bullet {(frac{e^{ix} - e^{-ix }}{2i})}^4 dx $$
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
$endgroup$
add a comment |
$begingroup$
You could also use Euler’s Formula.
$$ sin x = frac{e^{ix} - e^{-ix }}{2i} $$
$$ cos x = frac{e^{ix} + e^{-ix}}{2} $$
$$int {{(frac{e^{ix} - e^{-ix }}{2i})}^2} bullet {(frac{e^{ix} - e^{-ix }}{2i})}^4 dx $$
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
$endgroup$
You could also use Euler’s Formula.
$$ sin x = frac{e^{ix} - e^{-ix }}{2i} $$
$$ cos x = frac{e^{ix} + e^{-ix}}{2} $$
$$int {{(frac{e^{ix} - e^{-ix }}{2i})}^2} bullet {(frac{e^{ix} - e^{-ix }}{2i})}^4 dx $$
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
answered Feb 9 at 18:09
Michael LeeMichael Lee
6018
6018
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3084326%2fany-thoughts-on-this-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
As both exponents are even, you have to linearise. The simplest way uses the complex exponential.
$endgroup$
– Bernard
Jan 23 at 11:01
$begingroup$
You could try with $cos^2=1-sin^2$, then decompose $sin^4=sin^2cdotsin^2$, and similarly forma $sin^6$. Working this way you should lower the exponent by partial integration
$endgroup$
– Gabriele Cassese
Jan 23 at 11:01
$begingroup$
You can try t=sin^3(x)
$endgroup$
– Shaq
Jan 23 at 11:02